A child jumps on a trampoline, which of the following causes the child to rise in the air?

Which temperature scale does NOT have negative values?

german

Kelvin temperature scale

4 0

3 years ago

There is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted s

Maslowich

Answer:

.a = 849.05 m / s²

Explanation

The centripetal acceleration is

a = v² / r

Linear and angular velocity are related

v = w r

Angular velocity and frequency are related by

w = 2π f

Let's replace

a = w² r

a = 4π² f² r

Let's reduce to the SI system

f = 2.30 rev / s (2π rad / 1 rev) = 14.45 rad / s

.r = 10.3 cm = 0.103 m

Let's calculate

a = 4π² 14.45² 0.103

.a = 849.05 m / s²

8 0

3 years ago

Why is the overall charge of the atom neutral or zero?

MrRissso [65]

Answer:

B

Explanation:

this is because the neutrons do not have a charge, the things that have charge in an atom are electrons and protons.

and in the nucleus of an atom, there are protons and neutrons so you can see that A is not the answer

if you see the periodic table, you will know that the number of electrons and protons are equal, so the charges cancel each other out, hence the charge of an atom will be neutral

let me give you a tip which I got from my teacher, never write there is no charge in the atom, this suggests that there is no protons or electrons.

instead, write, the it is neutral

hope it helps if not please report it so that someone else gets to try it out

7 0

3 years ago

Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s

frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

y₁ = v₀₁ t + ½ g t²

y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

t ’= (t-t₀)

y₂ = v₀₂ t ’+ ½ g t’²

y₂ = 0 + ½ g (t-t₀)²

y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

S = y₁ -y₂

S = ½ g t²– ½ g (t-t₀)²

S = ½ g [t² - (t²- 2 t to + to²)]

S = ½ g (2 t t₀ - t₀²)

S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to. The rock y has not left and the distance increases

For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂

3 0

3 years ago

1. Bone has a Young’s modulus of about

Blababa [14]

#1

As we know that

$Y = \frac{stress}{strain}$

now plug in all data into this

$1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}$

$strain = 9.33 \times 10^{-3}$

now from the formula of strain

$strain = \frac{\Delta L}{L}$

$9.33 \times 10^{-3} = \frac{\Delta L}{0.54}$

$\Delta L = 5.04 \times 10^{-3} m$

$\Delta L = 5.04 mm$

#2

As we know that

pressure * area = Force

here we know that

$Area = 3.53 \times 11.6 = 40.95 m^2$

$P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa$

now force is given as

$F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N$

#3

As we know that density of water will vary with the height as given below

$\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}$

here we know that

$\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa$

$B = 2.3 \times 10^9 N/m^2$

now density is given as

$\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}$

$\rho = 1185.3 kg/m^3$

#4

as we know that pressure changes with depth as per following equation

$P = P_o + \rho g h$

here we know that

$P = 3 P_0$

now we will have

$3P_0 = P_0 + \rho g h$

$2P_0 = \rho g h$

$2(1.01 \times 10^5) = 1025 (9.81)(h)$

here we will have

$h = 20.1 m$

so it is 20.1 m below the surface

#5

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

$mg = \rho_1V_1g + \rho_2V_2g$

$A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)$

$46.6 = 43.89 - 922x + 1000x$

$x = 3.48 cm$

so it is 3.48 cm below the interface

5 0

3 years ago